I=∫[x^2/√(x^2+a^2)]dx =∫[(x^2+a^2-a^2)/√(x^2+a^2)]dx =∫√(x^2+a^2)dx-∫(a^2)/√(x^2+a^2)]dx =x√(x^2+a^2)-I-(a^2)ln[x+√(x^2+a^2)]+2C, 得 I=∫[x^2/√(x^2+a^2)]dx=(1/2)x√(x^2+a^2)-(a^2/2)ln[x+√(x^2+a^2)]+C
答：解：可用分部积分求出。设I=∫√(x²-1)dx，则 I=x√(x²-1)-∫xd√(x²-1) =x√(x²-1)-∫...详情>>
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